Assertion: Pyrrole is more basic than pyridine.

Reason: In pyrrole nitrogen is sp³ hybridized.

Both Assertion and Reason are incorrect.

The correct answer is option 4. Both Assertion and Reason are incorrect.

Assertion: Pyrrole is more basic than pyridine.

The assertion is incorrect. Pyridine is more basic than pyrrole. The nitrogen in pyridine is \(sp^2\) hybridized, and its lone pair of electrons is in an \( sp^2 \) orbital, which is not involved in the aromatic π-system of the ring. This makes the lone pair available for protonation, enhancing its basicity. Pyrrole is less basic compared to pyridine. In pyrrole, the nitrogen is also \(sp^2\) hybridized, but its lone pair of electrons is involved in the aromatic π-system. The involvement of the lone pair in the aromatic \(\pi \)-system delocalizes the electron density, making the lone pair less available for protonation. This delocalization decreases its basicity. Pyridine is more basic than pyrrole due to the availability of the lone pair on nitrogen in pyridine, which is not involved in the aromatic system.

Reason: In pyrrole, nitrogen is sp³ hybridized.

Pyrrole Hybridization: The nitrogen in pyrrole is \(sp^2\) hybridized. Pyrrole has a five-membered ring with four carbon atoms and one nitrogen atom. The nitrogen atom’s lone pair is in an \( sp^2 \) orbital, which is used to form the aromatic \(\pi \)-system with the carbon atoms in the ring. If the nitrogen were \(sp^3\) hybridized, its lone pair would be in a different orbital, and it would not be involved in the aromatic π-system. This incorrect hybridization does not reflect the actual electronic structure of pyrrole. The nitrogen in pyrrole is \(sp^2\) hybridized, not \(sp^3\). The incorrect hybridization description contributes to the incorrect reasoning about pyrrole's basicity.

Summary of Correct Understanding

Pyridine is more basic than pyrrole because the lone pair of electrons on nitrogen in pyridine is not involved in the aromatic π-system and is more available for protonation. Pyrrole’s nitrogen is \(sp^2\) hybridized, and its lone pair is part of the aromatic π-system, making it less available for protonation and thus less basic. Therefore, the assertion and reason provided are both incorrect